How do you find the circumference of the ellipse #x^2+4y^2=1#?

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Answer 1 · 2017-02-15T00:41:20

Using numerical techniques, we can get a approximation for this as:

# C = 4.8442 #

Although this seems like quite a simple question, the answer is actually ridiculous complicated.

We need to first put the ellipse equation in standard form:

#x^2+4y^2=1#
# :. (x/1)^2+(y/(1/2))^2=1#

Comparing with the standard equation;

# (x/a)^2+(y/b)^2=1#

We can identify this as an ellipse with semi-major axis #b=1/2# and semi-minor axis #a=1#, and the eccentricity of the ellipse is given by:

# e=sqrt(1-(b/a)^2)) #
# \ = sqrt(1-((1/2)/1)^2) #
# \ = sqrt(3/4) #
# \ = 1/2sqrt(3) #

Then the exact circumference is given by:

# C=4aE(e) #

where #E(e)# is a complete elliptic integral of the second kind,

# E(e) = int_0^(pi/2) \ sqrt(1-e^2sin^2 theta) \ d theta #

Using numerical techniques, we can get a approximation for this as:

# C = 4.8442 #